Question

obtain all other zeroes of the polynomial x4 -2x3-26x2+54x-27 if two of its zeroes are 3√3 and -3√3

Answer 1

x=3√3
⇒x-3√3=0
and,  x=-3√3
⇒x+3√3=0
g(x)=(x-3√3)(x+3√3)
      =x²-27
Now,
x²-27 ÷ x4 -2x3-26x2+54x-27
=x²-2x+1
now solving(x²-2x+1)
q(x)=x²-x-x+1
       =x(x-1)-1(x-1)
       =(x-1)(x-1)
Hence, other roots of the given p(x) are:   (x-1)



Hope it helped you!!...

Answer 2

Answer:

All Zeroes are -3√3 , 3√3 , 1 & 1.

Step-by-step explanation:

Given:  $p(x)=x^4-2x^3-26x^2+54x-27$

            Two zeroes are -3√3 & 3√3

To find: Other 2 Zeroes

We are given -3√3 & 3√3 as zeroes of p (x).

∴ expression made by zeroes are factors of p (x).

⇒ ( x + 2√3 ) & ( x - 3√3 )  are factors of p (x).

⇒ ( x + 3√3 ) × ( x - 3√3 ) = x² - 27 is also factor of p (x).

By dividing p (x) with x² - 27  we get another factor of p (x).

Long division is attached in pic.

⇒ p (x) = ( x² - 27 ) ( x² - 2x + 1 )

           = ( x + 3√3 )  ( x - 3√3 ) [ x² - x - x + 2 ]

           = ( x + 3√3 )  ( x - 3√3 ) [ x ( x - 1 ) - ( x - 1 ) ]

           = ( x + 3√3 )  ( x - 3√3 ) ( x - 1 ) ( x - 1 )

Zeroes, x - 1 = 0 ⇒ x = 1     and   x - 1 = 0 ⇒ x = 1

Therefore, All Zeroes are -3√3 , 3√3 , 1 & 1.